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Math Help - inequality

  1. #1
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    inequality

    when i have an inequality like this ,

    0<\sqrt{13}-3<1

    If i were to find (\sqrt{13}-3)^4 , can i say that

    0^4<(\sqrt{13}-3)^4<1^4 = 0<(\sqrt{13}-3)^4<1 ??

    And if the value at the side are negative , for instance

     <br />
-5<x<-1<br />
    If i were find range of x^2 , am i right in saying .

    25>x^2>1

    Rearrange 1<x^2<25
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  2. #2
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    Yes, you can
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  3. #3
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    Quote Originally Posted by songoku View Post
    Yes, you can

    THanks .

    How about only one of its side is negative ?

    -5<x<2

    TO find x^2 , do i do that this way :

    25>x^2>4

    Then rearrange 4<x^2<25
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  4. #4
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    Quote Originally Posted by thereddevils View Post
    THanks .

    How about only one of its side is negative ?

    -5<x<2

    TO find x^2 , do i do that this way :

    25>x^2>4

    Then rearrange 4<x^2<25
    x = 0 satisfies -5<x<2 . Does x = 0 satisfy 4<x^2<25 ....?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    x = 0 satisfies -5<x<2 . Does x = 0 satisfy 4<x^2<25 ....?

    THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
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  6. #6
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    Moral: do that as long as all pieces are positive.
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  7. #7
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    Quote Originally Posted by thereddevils View Post
    THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
    The range of f(x)= x^2 is the set of all non-negative numbers.
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  8. #8
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    Quote Originally Posted by thereddevils View Post
    How about only one of its side is negative ?

    -5<x<2

    TO find x^2 , do i do that this way :

    math]25>x^2>4 [/tex]

    Then rearrange 4<x^2<25
    Quote Originally Posted by mr fantastic View Post
    x = 0 satisfies -5<x<2 . Does x = 0 satisfy 4<x^2<25 ....?
    Quote Originally Posted by thereddevils View Post
    THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
    Draw a graph of y = x^2 for -5 < x < 2. Now look at the range of values of x^2.
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